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February 12

Is there are transition point between the spirals in the image when applying a Möbius transformation to spirals?

Hello,

this question has been bugging me for some time now. Let us consider a spiral in the complex plane, starting in zero and moving "towards infinity". A Möbius transformation will be a curve, acting like a spirak initially, and then at infinity approaching another point in the complex point, again like a spiral (but this time, it's turning in the other direction).

But... is there a "transition point"? A point where the curve stops being "like the first spiral" and becomes "the second spiral"? I've been trying to pinpoint this point, I tried to come up with definitions or even ways to calculate it, but the more and more I'm experimenting, the more I'm wondering whether or not there really is a point like that? Can anyone help? Thanks!

Evilbu (talk) 11:48, 12 February 2008 (UTC)[reply]

The transition point would be a point of inflexion, wouldn't it? They aren't usually hard to find. --Tango (talk) 13:40, 12 February 2008 (UTC)[reply]

Yeah. I think the choice of transition point has mostly to do with personal preference, but the inflection point (the point where the spiral changes direction of rotation) is probably the most obvious one. To find it, first find the equation of the transformed spiral, and then find the point where the ratio of the first and second derivatives has an imaginary part of 0. This will guarantee that the ratio is real, thus the second derivative (acceleration) has the same direction as the first (velocity), thus there is no rotation component and this is an inflection point. -- Meni Rosenfeld (talk) 14:14, 12 February 2008 (UTC)[reply]

I'm not so sure. I tried taking

\Gamma :[0,\infty )\rightarrow \mathbb {C} :r\rightarrow re^{ir}

and a Mobius transformation

M(z)={\dfrac {az+b}{z+d}}

and came out with the only point where the equation of the transformed spiral is 0 is

r=i

which lies outside the given range of r. -mattbuck (Talk) 23:24, 12 February 2008 (UTC)[reply]

I think you misunderstood. The transformed spiral probably won't go through 0, and the ratio of the second and first derivatives probably won't go through 0, but the ratio will cross the real axis. -- BenRG (talk) 02:21, 13 February 2008 (UTC)[reply]

I'll take this opportunity to mention that I wasn't completely accurate in my post. For a smooth curve, the criterion I have given is necessary, but not sufficient, for inflection. However, for the curve in question there will usually be only one point satisfying it, so it can be used to identify the point. -- Meni Rosenfeld (talk) 08:46, 13 February 2008 (UTC)[reply]

Oblique Asymptote

Is there a way to find the oblique asymptote of a polynomial function without using polynomial long division or calculus? Acceptable (talk) 23:52, 12 February 2008 (UTC)[reply]

Polynomials don't have asymptotes, do you mean a rational function? [1] has a good method - basically, you just take the limit of the function as x tends to infinity to find the gradient, and then take the limit of the function minus the bit you just found to find the constant term. --Tango (talk) 00:11, 13 February 2008 (UTC)[reply]

Sorry my bad, I indeed meant rational functions. Acceptable (talk) 01:23, 13 February 2008 (UTC)[reply]

There is actually a much simpler way. Just take the two highest terms of the numerator and divide them by the highest term of the denominator. For example, the asymptote of

{\frac {2x^{5}+3x^{3}-x^{2}+4x-5}{3x^{4}+8x^{3}-2x^{2}+5x-7}}

is

{\frac {2x^{5}+0x^{4}}{3x^{4}}}={\tfrac {2}{3}}x

. -- Meni Rosenfeld (talk) 08:34, 13 February 2008 (UTC)[reply]

Does that always work? It looks like a rough method of taking limits, but I think it only works on that function because the degree of the numerator is one higher than that of the denominator. You would need to adjust that method to work for other degrees. ~~You also still need to find the constant term.~~ --Tango (talk) 14:52, 13 February 2008 (UTC)[reply]

That's true. I did assume that the function satisfies this condition, as this is the usual case where the term "oblique asymptote" would be used. For finding the asymptote in a more general case, you need to take the terms of the numerator from the degree of the denominator and upwards. This still avoids limits, derivatives, or polynomial long division. -- Meni Rosenfeld (talk) 15:06, 13 February 2008 (UTC)[reply]

Don't think it works if the difference in degree is >1. e.g.

{\frac {x^{3}+2x^{2}+5x+4}{x+3}}

yields

{\frac {x^{3}+2x^{2}+5x}{x}}=x^{2}+3x+5

by your method. However, a quick long division yields

{\frac {x^{3}+2x^{2}+5x+4}{x+3}}=x^{2}-x+8-{\frac {20}{x+3}}

, i.e. an asymptote of x^2-x+8. Long division is the easiest method in such cases, methinks. --PalaceGuard008 (Talk) 22:19, 13 February 2008 (UTC)[reply]

You are of course absolutely right. Not one of my best days, apparently. -- Meni Rosenfeld (talk) 22:43, 13 February 2008 (UTC)[reply]